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I was reading some proofs that $X$ is a Hausdorff space $\iff$ the diagonal $\Delta = \{x\times x; x\in X\}$ is closed. I've found this one, that says:

Let X be Hausdorff, then if $x\ne y$ there are neighborhoods $V_x$ and $V_y$ such that $V_x \cap V_y = \emptyset$. Therefore $V_x\times V_y \cap D=\emptyset$ and the complement of $D$ is open. Now, assume that the latter is true. Then, for any point $(x,y)$, $x\ne y$, there is an open set around it that does not intersect $D$. Therefore, there are two sets $x\in V_x$ and $y\in V_y$ such that $V_x\times V_y$ doesn't intersect $D$, therefore $V_x \cap V_y = \emptyset$.

My first question is: in the second line, when he concludes that the complementar is open, why? For me, in order for a complement of $D$ to be open, I have to prove that $X-D$ is open. How to prove that? Also, is he considering $V_x\times V_y$ an open? Why? Because an open in the cartesian product is an union of $U\times V$, where $U$ is open in $X$ and $V$ is open in $X$.

Community
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Guerlando OCs
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  • The complement of $D$ is $(X \times X) \mathop{\backslash} D$. The text you cite proves that this is open. If $V_x$ and $V_y$ are open in $X$ then $V_x \times V_y$ is a union of sets of the form $U \times V$ where $U$ and $V$ are open in $X$. – Rob Arthan Sep 09 '16 at 23:26
  • You do not have to prove that $X-D$ is open, but that $X\times X-D$ is open. – drhab Sep 09 '16 at 23:29

2 Answers2

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Why is $V_x\times V_y$ open?

Because it is a union of sets of the form $U\times V$, where $U$ is open in $X$ and $V$ is open in $X$. Namely, it is the union of the single set $V_x\times V_y$, with $V_x$ and $V_y$ both open in $X$.

Why is the complement $X\times X - D$ open?

They are using the following lemma: $E$ is open if and only if, for all $x\in E$, there exists an open set $U$ such that $x\in U\subset E$.

Specifically, they showed that for all $(x,y)\in X\times X-D$ (meaning $x\neq y$), there exists an open set (namely, $V_x\times V_y$) such that $(x,y)\in V_x\times V_y\subset X\times X-D$. Note that saying $V_x\times V_y\subset X\times X-D$ is equivalent to saying $V_x\times V_y\cap D=\varnothing$.

Mike Earnest
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$\triangle\subseteq X\times X$ (and not $\triangle\subseteq X$).

For $\langle x,y\rangle\in\triangle^c= (X\times X)\setminus\triangle$ a neighborhood $V_x\times V_y$ is found that has empty intersection with $\triangle$. This means that $\langle x,y\rangle$ is not an element of the closure of $\triangle$.

It can be proved for every $\langle x,y\rangle\in\triangle^c$ so $\triangle$ contains its limit points hence is a closed subset of $X\times X$

$V_x$ and $V_y$ can both be chosen as open sets that contain $x$ resp. $y$ and in that case $V_x\times V_y$ is an open set of $X\times X$.

drhab
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